For the given initial-value problem y(0) = y'(0) = 0, solve for y(t) in an undamped spring-mass system with the specified parameters. Additionally, determine the maximum displacement of the mass, state if beats or resonance are present, and sketch the solution.
Given parameters:
m = 1
k = 25
f(t) = 0.01 \cos(5t)
Solution:
Solve the Homogeneous Equation
The corresponding homogeneous equation is:
y_h'' + 25y_h = 0
The characteristic equation is:
r^2 + 25 = 0 \quad \Rightarrow \quad r = \pm 5i
The general solution to the homogeneous equation is:
y_h(t) = C_1 \cos(5t) + C_2 \sin(5t)
Find the Particular Solution
The non-homogeneous term is f(t) = \frac{1}{100} \cos(5t). Since this term has the same frequency as the homogeneous solution, resonance occurs.
To account for resonance, assume a particular solution of the form:
y_p(t) = t \left( A \cos(5t) + B \sin(5t) \right)
Compute the Derivatives of y_p(t)
First Derivative y_p'(t):
Using the product rule:
\begin{aligned}
y_p'(t) &= A \cos(5t) + B \sin(5t) + t \left( -5A \sin(5t) + 5B \cos(5t) \right) \\
&= A \cos(5t) + B \sin(5t) - 5A t \sin(5t) + 5B t \cos(5t)
\end{aligned}
Second Derivative y_p''(t):
Differentiating y_p'(t):
\begin{aligned}
y_p''(t) &= -5A \sin(5t) + 5B \cos(5t) - 5A \sin(5t) + 5B \cos(5t) \\
&\quad - 25A t \cos(5t) - 25B t \sin(5t) \\
&= -10A \sin(5t) + 10B \cos(5t) - 25A t \cos(5t) - 25B t \sin(5t)
\end{aligned}
Substitute into the Original Equation
y_p''(t) + 25 y_p(t) = \frac{1}{100} \cos(5t)
Plugging in the expressions:
\begin{aligned}
&\left[ -10A \sin(5t) + 10B \cos(5t) - 25A t \cos(5t) - 25B t \sin(5t) \right] \\
&\quad + 25 \left[ A t \cos(5t) + B t \sin(5t) \right] = \frac{1}{100} \cos(5t)
\end{aligned}
Since resonance occurs, the amplitude of y(t) = \frac{t}{1000} \sin(5t) grows without bound as t \to \infty. So the maximum displacement increases indefinitely over time due to resonance.
Presence of Beats or Resonance
Resonance is present because the forcing frequency matches the natural frequency of the system.
Graph
Show Code
import numpy as npimport matplotlib.pyplot as pltt = np.linspace(0, 50, 1000)y = (t /1000) * np.sin(5* t)plt.figure(figsize=(10, 6))plt.plot(t, y, label=r"$y(t) = \frac{t}{1000} \sin(5t)$")plt.xlabel("Time (t)")plt.ylabel("Displacement (y)")plt.title("Graph of Solution")plt.legend()plt.grid(True)plt.show()
Final Answer:
y(t) = \frac{t}{1000} \sin(5t)
The system exhibits resonance, leading to an unbounded increase in displacement over time.
4.5.3
Problem:
In the given initial-value problem, solve for y(t) in an undamped spring-mass system with the specified parameters. Additionally, determine the maximum displacement of the mass, state if beats or resonance are present, and sketch the solution.
Given parameters:
m = 1
k = 36
f(t) = 2 e^{6t}
Initial conditions: y(0) = 0 and y'(0) = 0
Solution:
Solve the Homogeneous Equation
The homogeneous equation is:
y_h'' + 36 y_h = 0
The characteristic equation is:
r^2 + 36 = 0 \quad \Rightarrow \quad r = \pm 6i
The general solution to the homogeneous equation is:
y_h(t) = C_1 \cos(6t) + C_2 \sin(6t)
Find the Particular Solution
The non-homogeneous term is f(t) = 2 e^{6t}. Since this is an exponential function and does not overlap with the homogeneous solution, assume a particular solution of the form:
y_p(t) = A e^{6t}
Compute the Derivatives of y_p(t)
First Derivative y_p'(t):
y_p'(t) = 6 A e^{6t}
Second Derivative y_p''(t):
y_p''(t) = 36 A e^{6t}
Substitute into the Original Equation
Substitute y_p(t), y_p'(t), and y_p''(t) into the differential equation:
The term \frac{1}{36} e^{6t} grows exponentially as t \to \infty, leading to an unbounded increase in displacement. Therefore, the maximum displacement increases without bound over time.
Presence of Beats or Resonance
Resonance is not present in this system, as the forcing frequency does not match the natural frequency.
Beats are not present since there are no two competing frequencies close to each other.
The maximum displacement grows exponentially due to the \frac{1}{36} e^{6t} term.
SECTION 3.2
Assigned: 9, 11, 17
3.2.9
Problem:
Consider the system of differential equations given by
x_1' = -x_1 + 2x_2
x_2' = -7x_1 + 8x_2
Determine a matrix \mathbf{A} so that the system may be written in the form \mathbf{x}' = \mathbf{A} \mathbf{x}.
Determine all constant solutions to \mathbf{x}' = \mathbf{A} \mathbf{x}.
Compute the eigenvalues and eigenvectors of \mathbf{A}.
Determine all straight-line solutions to \mathbf{x}' = \mathbf{A} \mathbf{x}.
Find a more general solution to \mathbf{x}' = \mathbf{A} \mathbf{x} by taking all possible linear combinations of the straight-line solutions from (d).
Solve the initial-value problem \mathbf{x}' = \mathbf{A} \mathbf{x}, \mathbf{x}(0) = \begin{bmatrix} -2 \\ 0 \end{bmatrix}. Discuss the graphical behavior of this solution.
Graphical behavior: As t \to \infty, e^{6t} dominates e^{t}, so \mathbf{x}(t) grows exponentially in the direction of \begin{bmatrix} 2 \\ 7 \end{bmatrix}.
3.2.11
Problem:
Consider the system of differential equations given by
x_1' = -2x_1 + x_2
x_2' = -2x_2
Determine a matrix \mathbf{A} so that the system may be written in the form \mathbf{x}' = \mathbf{A} \mathbf{x}.
Determine all constant solutions to \mathbf{x}' = \mathbf{A} \mathbf{x}.
Compute the eigenvalues and eigenvectors of \mathbf{A}.
Determine all straight-line solutions to \mathbf{x}' = \mathbf{A} \mathbf{x}.
Find a more general solution to \mathbf{x}' = \mathbf{A} \mathbf{x} by taking all possible linear combinations of the straight-line solutions from (d).
Attempt to solve the initial-value problem \mathbf{x}' = \mathbf{A} \mathbf{x}, \mathbf{x}(0) = \begin{bmatrix} 1 \\ 1 \end{bmatrix}. What does this tell you about the proposed general solution in (e)?
Since \lambda = -2 is a repeated eigenvalue with only one linearly independent eigenvector, find a generalized eigenvector by solving (\mathbf{A} + 2\mathbf{I}) \mathbf{v} = \mathbf{v}_1:
The solution \mathbf{x}(t) = \begin{bmatrix} e^{-2t} \\ t e^{-2t} \end{bmatrix} shows that as t \to \infty, both components approach zero because of the e^{-2t} factor. However, the term t e^{-2t} causes the solution to decay slower in the x_2-direction, resulting in a solution that initially grows in the x_2-direction before eventually decaying to zero.
\mathbf{x}(t) = \begin{bmatrix} e^{-2t} \\ t e^{-2t} \end{bmatrix}
Interpretation: The term t e^{-2t} results in an initial growth in the x_2-direction before the solution eventually decays to zero, indicating a slower decay in the x_2-component due to the generalized eigenvector.
3.2.17
Problem:
Convert the second-order differential equation
y'' + y' - 12y = 0
into a system of first-order equations through an appropriate substitution.
Since W(\mathbf{x}_1, \mathbf{x}_2) = 3e^t, which is nonzero for all t, the vector functions \mathbf{x}_1(t) and \mathbf{x}_2(t) are linearly independent.
Final Answer:
The vector functions \mathbf{x}_1(t) and \mathbf{x}_2(t) are linearly independent because the Wronskian W(\mathbf{x}_1, \mathbf{x}_2) = 3e^t is nonzero for all t.
3.3.5
Problem:
Use the Wronskian to determine if the given set of vector functions is linearly independent:
\mathbf{x}_1(t) = \begin{bmatrix} \cos t \\ \sin t \end{bmatrix}, \quad \mathbf{x}_2(t) = \begin{bmatrix} \sin t \\ \cos t \end{bmatrix}
Solution:
To determine if \mathbf{x}_1(t) and \mathbf{x}_2(t) are linearly independent, calculate the Wronskian.
Set Up the Wronskian
The Wronskian W(\mathbf{x}_1, \mathbf{x}_2) of two vector functions \mathbf{x}_1(t) and \mathbf{x}_2(t) is defined as:
W(\mathbf{x}_1, \mathbf{x}_2) = \det \begin{bmatrix} \cos t & \sin t \\ \sin t & \cos t \end{bmatrix}
Calculate the Determinant
Expanding the determinant:
W(\mathbf{x}_1, \mathbf{x}_2) = \left( \cos t \cdot \cos t \right) - \left( \sin t \cdot \sin t \right)
= \cos^2 t - \sin^2 t
Using the trigonometric identity \cos^2 t - \sin^2 t = \cos(2t):
W(\mathbf{x}_1, \mathbf{x}_2) = \cos(2t)
Theorem 3.3.5
According to Theorem 3.3.5, a set of functions is linearly independent on an interval I if the Wronskian is nonzero at at least one point within I.
In this case, W(\mathbf{x}_1, \mathbf{x}_2) = \cos(2t), which is not identically zero on any open interval. Although \cos(2t) becomes zero at specific points (e.g., t = \frac{\pi}{4}, \frac{3\pi}{4}, \ldots), it is nonzero for many other points within any interval.
Final Answer:
Final Answer:
The vector functions \mathbf{x}_1(t) = \begin{bmatrix} \cos t \\ \sin t \end{bmatrix} and \mathbf{x}_2(t) = \begin{bmatrix} \sin t \\ \cos t \end{bmatrix} are linearly independent.
3.3.11
Problem:
Given the system \mathbf{x}' = A \mathbf{x} with matrix
A = \begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix}
Compute the eigenvalues and eigenvectors of A. Explain why you have found one linearly independent solution to the system, but still need to determine another.
Verify through direct substitution that \mathbf{x}_2(t) = t e^{3t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + e^{3t} \begin{bmatrix} 0 \\ 1 \end{bmatrix} is a solution to the given system \mathbf{x}' = A \mathbf{x}.
Show that the solution you found in (a) and the solution \mathbf{x}_2(t) in (b) are linearly independent, and hence state the general solution to the system.
Solve the initial value problem (IVP) with the initial condition \mathbf{x}(0) = \begin{bmatrix} 3 \\ 2 \end{bmatrix}.
Solution:
(a) Compute the Eigenvalues and Eigenvectors, and Explain the Need for Another Solution
Find the Eigenvalues:
To find the eigenvalues of A, solve \det(A - \lambda I) = 0:
A - \lambda I = \begin{bmatrix} 3 - \lambda & 1 \\ 0 & 3 - \lambda \end{bmatrix}
Solving \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}, we find that any vector of the form \mathbf{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} is an eigenvector. So there is only one linearly independent eigenvector \mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}.
Explain the Need for Another Solution:
According to Theorem 3.3.3, for an n \times n matrix A, we need n linearly independent eigenvectors to form a complete set of solutions for the system \mathbf{x}' = A \mathbf{x}. Each eigenvector \mathbf{v}_i with its corresponding eigenvalue \lambda_i provides a term c_i e^{\lambda_i t} \mathbf{v}_i in the general solution.
In our problem, the matrix A = \begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix} has a repeated eigenvalue \lambda = 3 but only one eigenvector. This is insufficient to meet the requirement of Theorem 3.3.3 for two linearly independent solutions (since A is 2 \times 2). Therefore, we need to find a second solution using a generalized eigenvector to form a complete, linearly independent solution set.
(b) Verify \mathbf{x}_2(t) = t e^{3t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + e^{3t} \begin{bmatrix} 0 \\ 1 \end{bmatrix} as a Solution
To verify that \mathbf{x}_2(t) is a solution, compute \mathbf{x}_2'(t) and check if \mathbf{x}_2'(t) = A \mathbf{x}_2(t).
Calculate \mathbf{x}_2'(t):
\mathbf{x}_2(t) = \begin{bmatrix} t e^{3t} \\ e^{3t} \end{bmatrix}
To show that \mathbf{x}_1(t) and \mathbf{x}_2(t) are linearly independent, consider a linear combination:
c_1 \mathbf{x}_1(t) + c_2 \mathbf{x}_2(t) = 0
Since \mathbf{x}_1(t) and \mathbf{x}_2(t) are not multiples of each other, they form a linearly independent set (you can verify this by calculating the Wronskian). So the general solution to the system is:
According to Theorem 3.3.3, to form a complete solution for the system \mathbf{x}' = A \mathbf{x}, we need two linearly independent solutions (since A is a 2 \times 2 matrix). However, since we only have one eigenvector, we must find a second solution by using a generalized eigenvector.
is a solution to the system \mathbf{x}' = A \mathbf{x}.
(c) The solutions \mathbf{x}_1(t) = e^{3t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} and \mathbf{x}_2(t) = t e^{3t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + e^{3t} \begin{bmatrix} 0 \\ 1 \end{bmatrix} are linearly independent. Therefore, the general solution to the system is:
(d) Solving the initial value problem with \mathbf{x}(0) = \begin{bmatrix} 3 \\ 2 \end{bmatrix}, we find c_1 = 3 and c_2 = 2. So the solution to the IVP is:
Let \mathbf{x}' = A \mathbf{x} be given by the matrix
A = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}
(a) Compute the eigenvalues and eigenvectors of A. Explain why, despite the repeated eigenvalue, two linearly independent solutions to the system are found.
(b) State the general solution to the system.
(c) Solve the initial value problem with the initial condition \mathbf{x}(0) = \begin{bmatrix} 3 \\ 2 \end{bmatrix}.
(d) Explain how the original system could be solved without using eigenvalues and eigenvectors.
Solution:
(a)Compute Eigenvalues and Eigenvectors
Finding the Eigenvalues:
To find the eigenvalues of A, solve the characteristic equation \det(A - \lambda I) = 0:
A - \lambda I = \begin{bmatrix} 3 - \lambda & 0 \\ 0 & 3 - \lambda \end{bmatrix}
This matrix has infinitely many solutions since any vector \mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} in \mathbb{R}^2 satisfies the equation. Two linearly independent vectors can be chosen, such as:
Despite having a repeated eigenvalue, two linearly independent solutions exist because the matrix A scales the standard basis vectors by the same factor (3). This means that any vector in \mathbb{R}^2 can be represented using the chosen eigenvectors.
(b)State the General Solution to the System
Given the two linearly independent eigenvectors, the general solution is:
This method solves the system directly by leveraging the diagonal structure of A, without requiring eigenvalues or eigenvectors.
Final Answer:
(a) The matrix A has an eigenvalue \lambda = 3, with two linearly independent eigenvectors \mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} and \mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}. The two solutions exist because A scales both basis vectors by the same factor, 3.
(d) Without using eigenvalues, the system can be solved by treating x_1 and x_2 as separate differential equations, leading to the same general solution:
where c_1 and c_2 are constants determined by initial conditions.
(b) The equilibrium solution at \mathbf{x} = \mathbf{0} is a saddle point and is unstable due to one positive and one negative eigenvalue.
(c) The straight-line solutions are along the lines x_2 = x_1 (unstable manifold) and x_2 = 3 x_1 (stable manifold). In the phase plane, trajectories exhibit saddle point behavior, moving away from the origin along the unstable direction and curving towards the stable direction when traced backward in time.
3.4.3
Problem:
Consider the system of differential equations \mathbf{x}' = A \mathbf{x} given by:
where c_1 and c_2 are constants determined by initial conditions.
(b) The equilibrium solution at \mathbf{x} = \mathbf{0} is a stable attracting node because both eigenvalues are negative.
(c) The straight-line solutions are along the lines x_2 = x_1 and x_2 = -x_1. In the phase plane, trajectories exhibit stable attracting node behavior,
with all paths approaching the origin.
3.4.7
Problem:
Consider the system of differential equations \mathbf{x}' = A \mathbf{x} with the general solution:
Since \lambda = -3 < 0, trajectories move toward the origin along this line as t \to \infty.
Part (b): Classify the Stability of the Equilibrium Solution
The equilibrium solution of the system \mathbf{x}' = A \mathbf{x} is at \mathbf{x} = \mathbf{0}.
Stability Analysis:
Eigenvalues:
\lambda_1 = 4
\lambda_2 = -3
Since one eigenvalue is positive and the other is negative, the equilibrium at the origin is classified as a saddle point.
Conclusion:
The equilibrium solution at \mathbf{x} = \mathbf{0} is unstable due to the presence of a positive eigenvalue.
Part (c): Sketch Straight-Line Solutions and Phase Plane Trajectories
Straight-Line Solutions:
Straight-line solutions correspond to trajectories along the eigenvectors.
Along \mathbf{v}_1 = \begin{bmatrix} -1 \\ 2 \end{bmatrix} (Unstable Eigenvector):
Trajectories move away from the origin as t \to \infty because e^{4t} grows exponentially.
Equation of the line: x_2 = -2x_1.
Along \mathbf{v}_2 = \begin{bmatrix} 1 \\ 2 \end{bmatrix} (Stable Eigenvector):
Trajectories approach the origin as t \to \infty because e^{-3t} decays exponentially.
Equation of the line: x_2 = 2x_1.
Graph
Show Code
# Corrected system of ODEs using the matrix Adef dx_dt(X, t): x1, x2 = X dx1 =0.5* x1 -1.75* x2 dx2 =-7* x1 +0.5* x2return [dx1, dx2]# Generate grid for direction fieldx1_vals = np.linspace(-10, 10, 40)x2_vals = np.linspace(-10, 10, 40)X1, X2 = np.meshgrid(x1_vals, x2_vals)# Calculate slopes for direction fieldU =0.5* X1 -1.75* X2V =-7* X1 +0.5* X2# Normalize the direction field arrowsspeed = np.sqrt(U**2+ V**2)epsilon =1e-10# Small value to prevent division by zeroU_norm = U / (speed + epsilon)V_norm = V / (speed + epsilon)# Plot the direction fieldplt.figure(figsize=(10, 10))plt.quiver(X1, X2, U_norm, V_norm, angles="xy", color="gray", scale=40, alpha=0.4, width=0.0035)# Plot the straight-line solutions (eigenvectors)x = np.linspace(-10, 10, 400)plt.plot(x, -2* x, 'r', linewidth=3, label=r'Unstable direction, $\lambda=4$')plt.plot(x, 2* x, 'b', linewidth=3, label=r'Stable direction, $\lambda=-3$')# Plot trajectories for different initial conditionsinitial_conditions = [ [2, 0], [-2, 0], [0, 5], [0, -5],]t_forward = np.linspace(0, 2, 500)t_backward = np.linspace(0, -2, 500)colors = ['green', 'purple', 'orange', 'cyan']for idx, ic inenumerate(initial_conditions):# Solve ODE forward in time sol_forward = odeint(dx_dt, ic, t_forward) plt.plot(sol_forward[:, 0], sol_forward[:, 1], color=colors[idx], linestyle='-', linewidth=2)# Solve ODE backward in time sol_backward = odeint(dx_dt, ic, t_backward) plt.plot(sol_backward[:, 0], sol_backward[:, 1], color=colors[idx], linestyle='-', linewidth=2)# Set plot limits and labelsplt.xlim(-10, 10)plt.ylim(-10, 10)plt.xlabel("$x_1$", fontsize=14)plt.ylabel("$x_2$", fontsize=14)plt.title(r"Direction Field and Trajectories for $\mathbf{x}' = A\mathbf{x}$", fontsize=16)# Add grid and legendplt.grid(True, linestyle='--', alpha=0.5)plt.legend(loc="upper left", fontsize=12, framealpha=1.0, frameon=True)# Display the plotplt.tight_layout()plt.show()
Final Answer:
(a) The straight-line solutions to the system are along the lines:
x_2 = -2x_1 (Unstable direction, \lambda = 4)
x_2 = 2x_1 (Stable direction, \lambda = -3)
(b) The equilibrium solution at \mathbf{x} = \mathbf{0} is classified as a saddle point and is unstable because one eigenvalue is positive.
(c) In the phase plane, trajectories exhibit saddle point behavior. Trajectories approach the origin along the line x_2 = 2x_1 (stable direction) and move away from the origin along x_2 = -2x_1 (unstable direction).
3.4.13
Problem:
Solve the initial value problem (IVP) for \mathbf{x}' = A \mathbf{x} with the given matrix A and initial condition.